Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{3x^2 + 30x}{x^2 - 100} \times \dfrac{x - 10}{x - 5} $
Explanation: First factor the quadratic. $n = \dfrac{3x^2 + 30x}{(x - 10)(x + 10)} \times \dfrac{x - 10}{x - 5} $ Then factor out any other terms. $n = \dfrac{3x(x + 10)}{(x - 10)(x + 10)} \times \dfrac{x - 10}{x - 5} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 3x(x + 10) \times (x - 10) } { (x - 10)(x + 10) \times (x - 5) } $ $n = \dfrac{ 3x(x + 10)(x - 10)}{ (x - 10)(x + 10)(x - 5)} $ Notice that $(x + 10)$ and $(x - 10)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 3x(x + 10)\cancel{(x - 10)}}{ \cancel{(x - 10)}(x + 10)(x - 5)} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $n = \dfrac{ 3x\cancel{(x + 10)}\cancel{(x - 10)}}{ \cancel{(x - 10)}\cancel{(x + 10)}(x - 5)} $ We are dividing by $x + 10$ , so $x + 10 \neq 0$ Therefore, $x \neq -10$ $n = \dfrac{3x}{x - 5} ; \space x \neq 10 ; \space x \neq -10 $